Question

How do you solve `root4(2x)+root4(x+3)=0`?

Answer 1

This equation has no solutions.

Explanation:

Note that `root(4)(t) >= 0` for all `t >= 0` and has no real value when `t < 0`.

So the only circumstances under which the left hand side of the given equation can be `0` are if both:

`2x=0" "` and `" "x+3=0`

The first of these implies that `x=0` and the second that `x=-3`.

These are incompatible, so there are no real solutions.

`color(white)()`
Complex solutions?

Given:

`root(4)(2x) + root(4)(x+3) = 0`

Subtract `root(4)(x+3)` from both sides to get:

`root(4)(2x) = -root(4)(x+3)`

Raise both sides to the `4`th power (noting that this is where we might introduce extraneous solutions) to get:

`2x = x+3`

Subtract `3` from both sides to find:

`x = 3`

But...

`root(4)(2(color(blue)(3)))+root(4)(color(blue)(3)+3) = root(4)(6)+root(4)(6) != 0`

So there are no solutions at all.